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Periodic energy space on an interval
PDF p. 2: §1.2, definition of H¹p(−π,π)
Definition
Assume the Axiom of Choice The Axiom of Choice. By The Axiom of Choice implies countable choice, this implies Countable Choice, as required by the Sobolev interfaces below. Let and write : its elements are complex equivalence classes whose first weak derivative also belongs to , equipped with the norm Integer-order Sobolev spaces and their norms. This expression is a well-defined norm by The Sobolev norm descends to equivalence classes, applied with , , and complex scalars. The periodic energy space is with the restricted norm. Here is the unique absolutely continuous representative on of the class ; its endpoint values are the Sobolev traces The trace of a one-dimensional Sobolev function is the pair of endpoint values.
Justification
By The trace of a one-dimensional Sobolev function is the pair of endpoint values, under the stated Axiom of Choice every has a uniquely determined absolutely continuous representative on the closed interval, and the endpoint trace map is linear and bounded. Therefore the equality of its endpoint values is independent of the representative of the class. Define by . Linearity of the trace gives for and . Thus is a linear subspace by The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial; the zero class belongs to it. The norm is well defined by restriction of the Sobolev norm in Integer-order Sobolev spaces and their norms, whose norm axioms and independence of representatives are supplied by The Sobolev norm descends to equivalence classes under Countable Choice, already obtained from the assumed Axiom of Choice.
Depends on
- The Axiom of Choice
- Integer-order Sobolev spaces and their norms
- The trace of a one-dimensional Sobolev function is the pair of endpoint values
- The Axiom of Choice implies countable choice
- The Sobolev norm descends to equivalence classes
- The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial
Dependency tree · 2 levels
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